codeforce 10B 思维题

http://vjudge.net/contest/view.action?cid=47807#problem/B

Description

All cinema halls in Berland are rectangles with K rows of K seats each, and K is an odd number. Rows and seats are numbered from 1 toK. For safety reasons people, who come to the box office to buy tickets, are not allowed to choose seats themselves. Formerly the choice was made by a cashier, but now this is the responsibility of a special seating program. It was found out that the large majority of Berland's inhabitants go to the cinema in order to watch a movie, that's why they want to sit as close to the hall center as possible. Moreover, a company of M people, who come to watch a movie, want necessarily to occupy M successive seats in one row. Let's formulate the algorithm, according to which the program chooses seats and sells tickets. As the request for M seats comes, the program should determine the row number x and the segment [y_l, y_r] of the seats numbers in this row, where y_r - y_l + 1 = M. From all such possible variants as a final result the program should choose the one with the minimum function value of total seats remoteness from the center. Say,  — the row and the seat numbers of the most "central" seat. Then the function value of seats remoteness from the hall center is . If the amount of minimum function values is more than one, the program should choose the one that is closer to the screen (i.e. the row number x is lower). If the variants are still multiple, it should choose the one with the minimum y_l. If you did not get yet, your task is to simulate the work of this program.

Input

The first line contains two integers N and K (1 ≤ N ≤ 1000, 1 ≤ K ≤ 99) — the amount of requests and the hall size respectively. The second line contains N space-separated integers M_i from the range [1, K] — requests to the program.

Output

Output N lines. In the i-th line output «-1» (without quotes), if it is impossible to find M_i successive seats in one row, otherwise output three numbers x, y_l, y_r. Separate the numbers with a space.

Sample Input

Input

2 1
1 1

Output

1 1 1
-1

Input

4 3
1 2 3 1

Output

2 2 2
1 1 2
3 1 3
2 1 1

题目大意：有一个k*k的电影院，给观众分配座位时，取  最小值，分配的座位必须在同一行且连续。若该最小值相等取i 较小的，还相等取j较小的。如果不能成功分配输出-1.若能，输出所在的行和所在列的起止点。

解题思路：范围0~10，暴力即可

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int mm=111;
int s[mm][mm],g[mm][mm];
int i,j,k,n,m;
int ab(int x)
{
    return x<0?-x:x;
}
void solve(int m)
{
    int i,j,x,y,tmp,ans=1e9;
    for(i=1;i<=k;++i)
    {
        for(j=1;j<=k;++j)
            if(!g[i][j])
            {
                if(j>=m&&s[i][j]-s[i][j-m]<ans)
                {
                    ans=s[i][j]-s[i][j-m];
                    x=i,y=j-m+1;
                }
            }
            else break;
        for(j=k-1;j>=0;--j)
            if(!g[i][j+1])
            {
                if(j+m<=k&&s[i][j+m]-s[i][j]<ans)
                {
                    ans=s[i][j+m]-s[i][j];
                    x=i,y=j+1;
                }
            }
            else break;
    }
    if(ans<1e9)
    {
        printf("%d %d %d\n",x,y,y+m-1);
        while(m--)g[x][y++]=1;
    }
    else puts("-1");
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(s,0,sizeof(s));
        memset(g,0,sizeof(g));
        for(i=1;i<=k;++i)
            for(j=1;j<=k;++j)
            {
                s[i][j]=ab(i-(k+1)/2)+ab(j-(k+1)/2);
                s[i][j]+=s[i][j-1];
            }
        while(n--)
        {
            scanf("%d",&m);
            solve(m);
        }
    }
    return 0;
}